Equipotential Surfaces: the Electric Potential Is Constant on the Surface • Equipotential Surfaces Are Perpendicular to Electric Field Lines

Physics 2102

Jonathan Dowling PPhhyyssicicss 22110022 LLeeccttuurree 55

EElleeccttrriicc PPootteennttiiaall II EElleeccttrriicc ppootteennttiiaall eenneerrggyy Electric potential energy of a system is equal to minus the work done by electrostatic forces when building the system (assuming charges were initially infinitely separated)

U= − W∞ The change in potential energy between an initial and final configuration is equal to minus the work done by the electrostatic forces:

ΔU= Uf − Ui= − W +Q • What is the potential energy of a single charge? +Q –Q • What is the potential energy of a dipole? a • A proton moves from point i to point f in a uniform electric field, as shown. • Does the electric field do positive or negative work on the proton? • Does the electric potential energy of the proton increase or decrease? EElleeccttrriicc ppootteennttiiaall

Electric potential difference between two points = work per unit charge needed to move a charge between the two points:

ΔV = Vf–Vi = −W/q = ΔU/q r dW = F • dsr r r dW = q0 E • ds f f r W = dW = q E • dsr # # 0 i i f W r !V = V "V = " = " E • dsr f i # q0 i EElleeccttrriicc ppootteennttiiaall eenneerrggyy,, eelleeccttrriicc ppootteennttiiaall Units : [U] = [W]=Joules; [V]=[W/q] = Joules/C= Nm/C= Volts [E]= N/C = Vm 1eV = work needed to move an electron through a potential difference of 1V: W=qΔV = e x 1V = 1.60 10–19 C x 1J/C = 1.60 10–19 J EEqquuipipootteennttiaiall ssuurrffaacceess f W r !V = V "V = " = " E • dsr f i q # Given a charged system, we can: 0 i • draw electric field lines: the electric field is tangent to the field lines • draw equipotential surfaces: the electric potential is constant on the surface • Equipotential surfaces are perpendicular to electric field lines. Why?? • No work is needed to move a charge along an equipotential surface. Why?? • Electric field lines always point towards equipotential surfaces with lower potential. Why?? ElectricElectric fieldfield lineslines andand eequipotentialquipotential ssurfacesurfaces

http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html EElleeccttrriicc ppootteennttiiaall aanndd eelleeccttrriicc ppootteennttiiaall eenneerrggyy The change in potential energy of a charge q moving from point i to point f is equal to the work done by the applied force, which is equal to minus the work done by the electric field, which is related to the difference in electric potential:

!U = U f "Ui = Wapp = "W = q!V We move a proton from point i to point f in a uniform electric field, as shown. • Does the electric field do positive or negative work on the proton? • Does the electric potential energy of the proton increase or decrease? • Does our force do positive or negative work ? • Does the proton move to a higher or lower potential? ExampleExample Consider a positive and a negative charge, freely moving in a uniform electric field. True or false? (a) Positive charge moves to points with lower potential. (b) Negative charge moves to points with lower potential. (c) Positive charge moves to a lower potential energy position. (d) Negative charge moves to a lower potential energy position (a) True +V (b) False –Q +Q 0 (c) True –V (d) True CCoonnsseerrvvaattivivee ffoorrcceess The potential difference between two points is independent of the path taken to calculate it: electric forces are “conservative”. f W !U r !V = V "V = " = = " E • dsr f i # q0 q0 i ElectricElectric PotentialPotential ofof aa PointPoint ChargeCharge

f r P V = "$ E #dsr = "$ E ds = i ! R kQ kQ R kQ = " dr = + = + $ 2 ! r r ! R

Note: if Q were a negative charge, V would be negative ElectricElectric PotentialPotential ofof ManyMany PointPoint ChargesCharges

• Electric potential is a q SCALAR not a vector. 4 q3 • Just calculate the potential r3 due to each individual r4 point charge, and add Pr together! (Make sure you 2 r5 q get the SIGNS correct!) q5 2 r1

qi q V = ! k 1 i ri EElleeccttrriicc ppootteennttiiaall aanndd eelleeccttrriicc ppootteennttiiaall eenneerrggyy

!U = Wapp = q!V +Q –Q a What is the potential energy of a dipole?

• First bring charge +Q: no work involved, no potential energy. • The charge +Q has created an electric potential everywhere, V(r)= kQ/r • The work needed to bring the charge –Q to a distance a from the charge +Q is 2 Wapp=U = (−Q)V = (–Q)(+kQ/a) = −kQ /a • The dipole has a negative potential energy equal to −kQ2/a: we had to do negative work to build the dipole (and the electric field did positive work). PotentialPotential EnergyEnergy ofof AA SystemSystem ofof ChargesCharges • 4 point charges (each +Q and equal mass) are connected by strings, forming a square of side L • If all four strings suddenly snap, +Q +Q what is the kinetic energy of each charge when they are very far apart? • Use conservation of energy: – Final kinetic energy of all four charges = initial potential energy stored = +Q +Q energy required to assemble the system of charges

Do this from scratch! PotentialPotential EnergyEnergy ofof AA SystemSystem ofof Charges:Charges: SolutionSolution L +Q +Q • No energy needed to bring in first charge: U =0 1 2L • Energy needed to bring in kQ2 2nd charge: U = QV = 2 1 L +Q +Q • Energy needed to bring in 3rd charge = Total potential energy is sum of all the individual terms shown kQ2 kQ2 2 U = QV = Q(V +V ) = + on left hand side = kQ 3 1 2 L 2L (4 + 2) L • Energy needed to bring in So, final kinetic energy of each 4th charge = charge = kQ2 2kQ2 kQ2 U = QV = Q(V +V +V ) = + (4 + 2) 4 1 2 3 L 2L 4L SSuummmmaarryy:: • Electric potential: work needed to bring +1C from infinity; units V = Volt • Electric potential uniquely defined for every point in space - - independent of path! • Electric potential is a scalar — add contributions from individual point charges • We calculated the electric potential produced by a single charge: V=kq/r, and by continuous charge distributions : V=∫ kdq/r • Electric potential energy: work used to build the system, charge by charge. Use W=qV for each charge.